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Find expectation $Eexp(X_1+X_2)mathbf{I}_{{X_1

0

$begingroup$

Let $X_1$ and $X_2$ two independent random variables. The PDF of $X_i$ is $f_{X_i}(x_i)$.

Ho we can find the following expectation
$$Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}.$$

Is it

begin{align}
Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}&=Ee^{X_1}e^{X_2}mathbf{I}_{{X_1<X_2}}\
&=int_{x_2=-infty}^{infty}e^{x_2}f_{X_2}(x_2)left(int_{x_1=-infty}^{x_2}e^{x_1}f_{X_1}(x_1)dx_1right)dx_2.
end{align}
Thanks.

expected-value

share|cite|improve this question

edited Mar 26 at 1:18

Monir

asked Mar 25 at 23:54

MonirMonir

539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does $EZ[A]$ mean when $Z$ is a random variable and $A$ is an event?.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Believe me I do not know.
  $endgroup$
  – Monir
  Mar 26 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not mean $E[Z|A]$. I mean the expected value of product of two random variables, $X_1$ and $X_2$ where $X_1<X_2$.
  $endgroup$
  – Monir
  Mar 26 at 0:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $$Eleft[X_1times X_2 [X_1<X_2]right]=int_{x_2=-infty}^{infty}x_2f_{X_2}(x_2)int_{x_1=-infty}^{x_2}x_1f_{X_1}(x_1)dx_1dx_2$$?
  $endgroup$
  – Monir
  Mar 26 at 0:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think that is correct but the notation used in the question is not satndard.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 0:18
  

  
  
  
  

 | 
show 6 more comments

0

$begingroup$

Let $X_1$ and $X_2$ two independent random variables. The PDF of $X_i$ is $f_{X_i}(x_i)$.

Ho we can find the following expectation
$$Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}.$$

Is it

begin{align}
Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}&=Ee^{X_1}e^{X_2}mathbf{I}_{{X_1<X_2}}\
&=int_{x_2=-infty}^{infty}e^{x_2}f_{X_2}(x_2)left(int_{x_1=-infty}^{x_2}e^{x_1}f_{X_1}(x_1)dx_1right)dx_2.
end{align}
Thanks.

expected-value

share|cite|improve this question

edited Mar 26 at 1:18

Monir

asked Mar 25 at 23:54

MonirMonir

539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does $EZ[A]$ mean when $Z$ is a random variable and $A$ is an event?.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Believe me I do not know.
  $endgroup$
  – Monir
  Mar 26 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not mean $E[Z|A]$. I mean the expected value of product of two random variables, $X_1$ and $X_2$ where $X_1<X_2$.
  $endgroup$
  – Monir
  Mar 26 at 0:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $$Eleft[X_1times X_2 [X_1<X_2]right]=int_{x_2=-infty}^{infty}x_2f_{X_2}(x_2)int_{x_1=-infty}^{x_2}x_1f_{X_1}(x_1)dx_1dx_2$$?
  $endgroup$
  – Monir
  Mar 26 at 0:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think that is correct but the notation used in the question is not satndard.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 0:18
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

Let $X_1$ and $X_2$ two independent random variables. The PDF of $X_i$ is $f_{X_i}(x_i)$.

Ho we can find the following expectation
$$Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}.$$

Is it

begin{align}
Eexp(X_1+X_2)mathbf{I}_{{X_1<X_2}}&=Ee^{X_1}e^{X_2}mathbf{I}_{{X_1<X_2}}\
&=int_{x_2=-infty}^{infty}e^{x_2}f_{X_2}(x_2)left(int_{x_1=-infty}^{x_2}e^{x_1}f_{X_1}(x_1)dx_1right)dx_2.
end{align}
Thanks.

expected-value

share|cite|improve this question

edited Mar 26 at 1:18

Monir

asked Mar 25 at 23:54

MonirMonir

539

$endgroup$

share|cite|improve this question

edited Mar 26 at 1:18

Monir

asked Mar 25 at 23:54

MonirMonir

539

share|cite|improve this question

edited Mar 26 at 1:18

Monir

asked Mar 25 at 23:54

MonirMonir

539

asked Mar 25 at 23:54

MonirMonir

539

asked Mar 25 at 23:54

MonirMonir

539